To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:
Where,
P = Axial Load
l = Gage length
A = Cross-sectional Area
E = Modulus of Elasticity
Our values are given as,
l = 3.5m
D = 0.028m
E = 200GPa
Replacing we have,
Therefore the change in length is 1.93mm
The magnitude and direction of the velocity of the car at t = 8.00 s is v = 7,468 m / s and the vector direction = θ = 58,984°
<h3><em>Further explanation</em></h3>In the coordinates of the two-dimensional plane (xy) there is a position vector of a point
You can write P as:
P = xi + yi
where i denotes the unit vector in the x and y directions
While the magnitude of the vector:
and its direction
Likewise with the instantaneous speed which is a derivative of the position function of time
V = vx i + vy j
The velocity of the car as a function of time is given by v = [5.00 m / s - (0.0180 m / s²) t²] i + [2.00 m / s + (0.550 m / s²) t] j so that:
vx = 5 - 0.0180t² i
vy = 2 + 0.55t j
So the instantaneous velocity at t = 8 s becomes:
vx = 5-0.0180.8²
vx = 3,848 m / s
vy = 2+ 0.55.8
vy = 6.4
So the velocity position vector becomes:
v = 3,848 i + 6.4 y
and the magnitude becomes:
v = 7,468 m / s
While the velocity vector direction:
θ = 58,984
<h3><em>Learn more</em></h3>Which vector should be negative
the statements that describe a vector
the resultant vector
Keywords: velocity, vector, two-dimensional coordinates,
Explanation:
Below is an attachment containing the solution
